Given,
∣x2−8x+15∣−2x+7=0
Now taking Case:I when,
x2−8x+15≥0⇒(x−3)(x−5)≥0
⇒x≤3 OR x≥5
∣x2−8x+15∣−2x+7=0
⇒x2−8x+15−2x+7=0
⇒x2−10x+22=0
⇒x=5+3,5−3(rejected)
So, α=5+3
Now taking Case:II when,
∣x2−8x+15∣−2x+7=0
⇒x2−8x+15<0⇒3<x<5
⇒x2−8x+15+2x−7=0
⇒x2−6x+8=0⇒x=4,2(rejected)
So, γ=4
Hence, α+γ=9+3